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4. The critical magnetic field The critical current density and the critical magnetic field The supercurrent density has a limit: JC When the superconductor is applied with a magnetic field, a supercurrent is generated so as to maintain the perfect diamagnetism. If the current density needed to screen the field exceeds JC, the superconductor will lose its superconductivity. This limit of the field strength is called the critical magnetic field HC. Note the difference between the flux density, B and field strength, H Free energy The magnetisation depends on the applied field and temperature but not the history. The transition is a reversible process the Gibbs free energy g s (T , H a ) is a function of temperature and field. Why Gibbs free energy? Because the system is in contact with a thermal reservoir, (T=constant) a pressure reservoir (P=constant), and a “magnetic-field reservoir”(H=constant) In this case, the system will minimize the Gibbs free energy Below TC The free energy of the superconducting state < The free energy of the normal state g s (T , 0 ) < g n (T , 0 ) The free energy increases when applied a magnetic field: ∆g ( H a ) = − µ0 Ha ∫ MdH a 0 Area of the M-H curve Here we neglect the effect of demagnetizing factor Free energy g s (T , H a ) = g s (T , 0 ) − µ0 g s (T , H a ) Ha ∫ MdH a g n (T , 0 ) g n (T , H a ) 0 = g s ( T , 0 ) + µ0 Ha ∫ HdH 0 a g s (T , 0 ) H a2 = g s ( T , 0 ) + µ0 2 HC H The normal state is non-magnetic and the applied field does not change the free energy much. When g s (T , H ) > g n (T , H ) the superconductivity is destryed . H C2 µ0 = g n (T , 0 ) − g s (T , 0 ) 2 The temperature dependence Phase diagram H Normal state H0 HC SC TC Experimental result: T ⎡ ⎛ T ⎞2 ⎤ H C = H 0 ⎢1 − ⎜ ⎟ ⎥ = H 0 (1 − t 2 ) ⎢⎣ ⎝ TC ⎠ ⎥⎦ The cryotron Gate(Ta) Current I When iC is applied, the field may destroy the superconductivity of the gate and reduce the current I iC Control(Nb) An electronic switch The M-H curve Deep inside the superconductor M B HC HC H area: the free energy contributed by M For a perfect specimen, the curves are reversible H The hysteresis: non-ideal specimens Residual magnetization: trapped flux • Ill defined HC • Hysteresis • Trapped flux Measurement of B C: pick-up coil S: solenoid to apply the field When the switch is closed, the ballistic galvanometer is deflected. The deflection of G is proportional to B. LR circuit Ta at 3.7K Measurement of magnetization VSM The specimen is moved to and fro between pick-up coils A and B. The galvanometer swings with an amount proportional to the magnetization M. Integrating method E = E A − EB d d ∝ µ0 ( H + M ) − µ0 H dt dt dM E∝ dt Integrator: Vout ∝ ∫ Vin dt ∝ M cf: AC susceptometer Homework 1. Use the definition of Gibbs free energy G = U − TS + PV − µ0 H a M to show the Gibbs free energy change due to a magnetic field Ha is given by. H ∆g ( H a ) = − µ0 ∫ MdH a a 0 5. Thermodynamics Entropy From We have H C2 µ0 = g n (T , 0 ) − g s (T , 0 ) 2 µ0 H ( 2 2 C − H a2 ) = g n (T , 0 ) − g s (T , H a ) G = U − TS + PV − µ0 H a M dG = dU − d (TS ) + d ( PV ) − µ0 d ( H a M ) = − SdT + VdP − µ0 MdH a entropy thus since ⎛ ∂g ⎞ s = −⎜ ⎟ ∂ T ⎝ ⎠ p,Ha HC is temperature dependent dH C sn − s s = µ 0 H C dT dH C <0 dT sn > s s The superconducting state is more “order” than the normal state H Normal state H0 HC SC TC T At TC , HC=0 sn = s s According to the third law of thermodynamics, sn = s s sn = s s at T=0 Since H0 is finite, this requires that dH C =0 dT at T=0 Entropy vs temperature 2nd order phase transitions At TC sn = s s ⎛ ∂g ⎞ ⎛ ∂g ⎞ ⎜ ⎟ =⎜ ⎟ T T ∂ ∂ ⎝ ⎠n ⎝ ⎠s ∂g Both g and its derivative are continuous ∂T A 2nd order transition Features: no latent heat L = vT ( ss − sn ) a jump in specific heat ∂s C = vT ∂T Specific heat 2 ⎡ ∂ ( s s − sn ) d 2 H C ⎛ dH C ⎞ ⎤ Cs − Cn = vT = vT µ0 ⎢ H C +⎜ ⎟ ⎥ ∂T dT dT ⎝ ⎠ ⎥⎦ ⎢⎣ At TC , HC=0 ( Cs − Cn )T C Jump at TC 2 ⎛ dH C ⎞ = vTC µ0 ⎜ ⎟ ⎝ dT ⎠TC Rutger’s formula Latent heat at non-zero field dH C L = vT ( ss − sn ) = −vT µ0 H C dT H H0 Normal state sn > s s 1st order transition SC 2nd order transition TC T Adiabatic magnetization Adiabatic process, s=constant dQ = TdS = 0 Adiabatically destructing the superconductivity by applying a large field can lower the temperature Adiabatic demagnetization Is used for ordinary magnetic materials in which the entropy decreases with application of field. Specific heat The specific heat of a metal is contributed from the lattice and conduction electrons C = Clatt + Cel The properties of the lattice do not change at the transition Cs − Cn = ( Cel ) s − ( Cel )n For a normal metal Cn = Clatt + ( Cel )n θ: Debye temperature 3 ⎛T ⎞ = A⎜ ⎟ + γ T ⎝θ ⎠ γ: Sommerfeld constant At T ~TC C s > Cn because sn > ss dS s dS n > dT dT At low temperature C s < Cn Cn A 2 = 3 T +γ T θ C/T At T<TC , one can measure C for normal state (Ha>HC) The electronic part for superconducting state can be determined − b / kT C = ae ( el )s Slope=A/θ γ TC T2 The form of the thermal activation : hint of a energy gap b/e As T approaches TC, b rapidly decreases to zero Irreversible processes Thermal conduction The thermal conduction in a metal is mostly contributed by the conduction electrons Superelectrons have negligible interaction with the lattice The thermal conductivity in superconducting state is much smaller than that in normal state. Thermoelectric effects Due to zero resistance of the superconductors, the thermal e.m.f. should be zero. Therefore, Peltier and Thomson coefficients are zero. In fact, thermoelectric effects may appear in type-II superconductors Seebeck(thermoelectric) effect e.m.f. due to temperature gradient under conditions of zero electric current A T1 B Peltier effect Heat flow following an electric current across an isothermal junction A V B current Heat flow Thomson effect Electric current traverses a temperature gradient T2 T1 T2 current Homework 1. Apply Sommerfeld’s theory to show that the electronic specific heat of the metals is proportional to the temperature, i. e. Cel = γ T in which the constant γ is proportional to the density of states at Fermi surface. 2. Do the same calculation by assuming an energy gap ∆ forming at the Fermi surface and show that Cel = ae − e∆ / kT at low temperature, kT<<∆